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Question

A coil with \( L = 0.15 \, \text{H} \) has its current increased from 0 to 3 A in 0.3 s. What is the
energy stored?

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Explanation

**Energy stored in inductor** U =½ L I² (J), L inductance (H), I current (A), resides in magnetic field, energy density u = B²/(2μ₀) (J/m³), B field (T), μ₀=4π×10⁻⁷ H/m. For solenoid B=μ₀ n I, volume V= A l, U = (B²/2μ₀) V =½ (μ₀ n² A l) I² =½ L I², consistent. W = (1/2) L I² = (1/2) × 0.15 × (3)² = 0.075 × 9 = 0.675 J . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U

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